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Calculus 3 : Partial Derivatives

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Example Questions

Example Question #41 : Partial Derivatives

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow10+}\frac{(10x - 100)}{(19x^{2} - 323x + 1330)}\end{align*}$ $\displaystyle \begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow10+}\frac{(10x - 100)}{(19x^{2} - 323x + 1330)}\end{align*}$

Possible Answers:

$\frac{10}{57}$ $\displaystyle \frac{10}{57}$

$-\frac{10}{57}$ $\displaystyle -\frac{10}{57}$

$-\infty$ $\displaystyle -\infty$

$\infty$ $\displaystyle \infty$

Correct answer:

$\frac{10}{57}$ $\displaystyle \frac{10}{57}$

Explanation:

$\begin{align*}&\text{We cannot simply plug in the value of }x=10\text{, as that would}\\&\text{create a zero value in the denominator, and not tell us much as is.}\\&\text{However, note that the expression}\\&\frac{(10x - 100)}{(19x^{2} - 323x + 1330)}\\&\text{can be factored. This will allow us to find the limit}\\&\text{taken from the right:}\\&\frac{(10\cdot (x - 10))}{(19\cdot (x - 10)\cdot (x - 7))}\\&\frac{10}{(19\cdot (x - 7))}\\&lim_{x\rightarrow10+}\frac{(10x - 100)}{(19x^{2} - 323x + 1330)}=\frac{10}{57}\end{align*}$ $\displaystyle \begin{align*}&\text{We cannot simply plug in the value of }x=10\text{, as that would}\\&\text{create a zero value in the denominator, and not tell us much as is.}\\&\text{However, note that the expression}\\&\frac{(10x - 100)}{(19x^{2} - 323x + 1330)}\\&\text{can be factored. This will allow us to find the limit}\\&\text{taken from the right:}\\&\frac{(10\cdot (x - 10))}{(19\cdot (x - 10)\cdot (x - 7))}\\&\frac{10}{(19\cdot (x - 7))}\\&lim_{x\rightarrow10+}\frac{(10x - 100)}{(19x^{2} - 323x + 1330)}=\frac{10}{57}\end{align*}$

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Example Question #41 : Limits

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow-6-}-\frac{(99x + 9x^{2} + 270)}{(663x - 104x^{2} - 13x^{3} + 4914)}\end{align*}$ $\displaystyle \begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow-6-}-\frac{(99x + 9x^{2} + 270)}{(663x - 104x^{2} - 13x^{3} + 4914)}\end{align*}$

Possible Answers:

$\displaystyle 0$

$-\infty$ $\displaystyle -\infty$

$\infty$ $\displaystyle \infty$

$\frac{3}{169}$ $\displaystyle \frac{3}{169}$

Correct answer:

$\frac{3}{169}$ $\displaystyle \frac{3}{169}$

Explanation:

$\begin{align*}&\text{We cannot simply plug in the value of }x=-6\text{, as that would}\\&\text{create a zero value in the denominator, and not tell us much as is.}\\&\text{However, note that the expression}\\&-\frac{(99x + 9x^{2} + 270)}{(663x - 104x^{2} - 13x^{3} + 4914)}\\&\text{can be factored. This will allow us to find the limit}\\&\text{taken from the left:}\\&\frac{(9\cdot (x + 6)\cdot (x + 5))}{(13\cdot (x + 6)\cdot (x - 7)\cdot (x + 9))}\\&\frac{(9\cdot (x + 5))}{(13\cdot (x - 7)\cdot (x + 9))}\\&lim_{x\rightarrow-6-}-\frac{(99x + 9x^{2} + 270)}{(663x - 104x^{2} - 13x^{3} + 4914)}=\frac{3}{169}\end{align*}$ $\displaystyle \begin{align*}&\text{We cannot simply plug in the value of }x=-6\text{, as that would}\\&\text{create a zero value in the denominator, and not tell us much as is.}\\&\text{However, note that the expression}\\&-\frac{(99x + 9x^{2} + 270)}{(663x - 104x^{2} - 13x^{3} + 4914)}\\&\text{can be factored. This will allow us to find the limit}\\&\text{taken from the left:}\\&\frac{(9\cdot (x + 6)\cdot (x + 5))}{(13\cdot (x + 6)\cdot (x - 7)\cdot (x + 9))}\\&\frac{(9\cdot (x + 5))}{(13\cdot (x - 7)\cdot (x + 9))}\\&lim_{x\rightarrow-6-}-\frac{(99x + 9x^{2} + 270)}{(663x - 104x^{2} - 13x^{3} + 4914)}=\frac{3}{169}\end{align*}$

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Example Question #41 : Limits

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow-2-}\frac{(1508x + 260x^{2} + 13x^{3} + 2080)}{(156x + 54x^{2} + 6x^{3} + 144)}\end{align*}$ $\displaystyle \begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow-2-}\frac{(1508x + 260x^{2} + 13x^{3} + 2080)}{(156x + 54x^{2} + 6x^{3} + 144)}\end{align*}$

Possible Answers:

$\infty$ $\displaystyle \infty$

$\displaystyle 0$

$\displaystyle 52$

$-\infty$ $\displaystyle -\infty$

Correct answer:

$\displaystyle 52$

Explanation:

$\begin{align*}&\text{We cannot simply plug in the value of }x=-2\text{, as that would}\\&\text{create a zero value in the denominator, and not tell us much as is.}\\&\text{However, note that the expression}\\&\frac{(1508x + 260x^{2} + 13x^{3} + 2080)}{(156x + 54x^{2} + 6x^{3} + 144)}\\&\text{can be factored. This will allow us to find the limit}\\&\text{taken from the left:}\\&\frac{(13\cdot (x + 2)\cdot (x + 8)\cdot (x + 10))}{(6\cdot (x + 2)\cdot (x + 3)\cdot (x + 4))}\\&\frac{(13\cdot (x + 8)\cdot (x + 10))}{(6\cdot (x + 3)\cdot (x + 4))}\\&lim_{x\rightarrow-2-}\frac{(1508x + 260x^{2} + 13x^{3} + 2080)}{(156x + 54x^{2} + 6x^{3} + 144)}=52\end{align*}$ $\displaystyle \begin{align*}&\text{We cannot simply plug in the value of }x=-2\text{, as that would}\\&\text{create a zero value in the denominator, and not tell us much as is.}\\&\text{However, note that the expression}\\&\frac{(1508x + 260x^{2} + 13x^{3} + 2080)}{(156x + 54x^{2} + 6x^{3} + 144)}\\&\text{can be factored. This will allow us to find the limit}\\&\text{taken from the left:}\\&\frac{(13\cdot (x + 2)\cdot (x + 8)\cdot (x + 10))}{(6\cdot (x + 2)\cdot (x + 3)\cdot (x + 4))}\\&\frac{(13\cdot (x + 8)\cdot (x + 10))}{(6\cdot (x + 3)\cdot (x + 4))}\\&lim_{x\rightarrow-2-}\frac{(1508x + 260x^{2} + 13x^{3} + 2080)}{(156x + 54x^{2} + 6x^{3} + 144)}=52\end{align*}$

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Example Question #1002 : Calculus 3

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow-5+}\frac{(18x + 90)}{(81x + 9x^{2} + 180)}\end{align*}$ $\displaystyle \begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow-5+}\frac{(18x + 90)}{(81x + 9x^{2} + 180)}\end{align*}$

Possible Answers:

$\infty$ $\displaystyle \infty$

$\displaystyle 2$

$-\infty$ $\displaystyle -\infty$

$\displaystyle -2$

Correct answer:

$\displaystyle -2$

Explanation:

$\begin{align*}&\text{We cannot simply plug in the value of }x=-5\text{, as that would}\\&\text{create a zero value in the denominator, and not tell us much as is.}\\&\text{However, note that the expression}\\&\frac{(18x + 90)}{(81x + 9x^{2} + 180)}\\&\text{can be factored. This will allow us to find the limit}\\&\text{taken from the right:}\\&\frac{(18\cdot (x + 5))}{(9\cdot (x + 5)\cdot (x + 4))}\\&\frac{2}{(x + 4)}\\&lim_{x\rightarrow-5+}\frac{(18x + 90)}{(81x + 9x^{2} + 180)}=-2\end{align*}$ $\displaystyle \begin{align*}&\text{We cannot simply plug in the value of }x=-5\text{, as that would}\\&\text{create a zero value in the denominator, and not tell us much as is.}\\&\text{However, note that the expression}\\&\frac{(18x + 90)}{(81x + 9x^{2} + 180)}\\&\text{can be factored. This will allow us to find the limit}\\&\text{taken from the right:}\\&\frac{(18\cdot (x + 5))}{(9\cdot (x + 5)\cdot (x + 4))}\\&\frac{2}{(x + 4)}\\&lim_{x\rightarrow-5+}\frac{(18x + 90)}{(81x + 9x^{2} + 180)}=-2\end{align*}$

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Example Question #1001 : Calculus 3

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow-4+}\frac{(45x + 5x^{2} + 100)}{(5x + 20)}\end{align*}$ $\displaystyle \begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow-4+}\frac{(45x + 5x^{2} + 100)}{(5x + 20)}\end{align*}$

Possible Answers:

$\displaystyle 1$

$\infty$ $\displaystyle \infty$

$-\infty$ $\displaystyle -\infty$

$\displaystyle 0$

Correct answer:

$\displaystyle 1$

Explanation:

$\begin{align*}&\text{We cannot simply plug in the value of }x=-4\text{, as that would}\\&\text{create a zero value in the denominator, and not tell us much as is.}\\&\text{However, note that the expression}\\&\frac{(45x + 5x^{2} + 100)}{(5x + 20)}\\&\text{can be factored. This will allow us to find the limit}\\&\text{taken from the right:}\\&\frac{(5\cdot (x + 4)\cdot (x + 5))}{(5\cdot (x + 4))}\\&x + 5\\&lim_{x\rightarrow-4+}\frac{(45x + 5x^{2} + 100)}{(5x + 20)}=1\end{align*}$ $\displaystyle \begin{align*}&\text{We cannot simply plug in the value of }x=-4\text{, as that would}\\&\text{create a zero value in the denominator, and not tell us much as is.}\\&\text{However, note that the expression}\\&\frac{(45x + 5x^{2} + 100)}{(5x + 20)}\\&\text{can be factored. This will allow us to find the limit}\\&\text{taken from the right:}\\&\frac{(5\cdot (x + 4)\cdot (x + 5))}{(5\cdot (x + 4))}\\&x + 5\\&lim_{x\rightarrow-4+}\frac{(45x + 5x^{2} + 100)}{(5x + 20)}=1\end{align*}$

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Example Question #43 : Partial Derivatives

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow7+}-\frac{(183x + 12x^{2} - 3x^{3} - 840)}{(19x - 133)}\end{align*}$ $\displaystyle \begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow7+}-\frac{(183x + 12x^{2} - 3x^{3} - 840)}{(19x - 133)}\end{align*}$

Possible Answers:

$\frac{90}{19}$ $\displaystyle \frac{90}{19}$

$\displaystyle 0$

$-\infty$ $\displaystyle -\infty$

$\infty$ $\displaystyle \infty$

Correct answer:

$\frac{90}{19}$ $\displaystyle \frac{90}{19}$

Explanation:

$\begin{align*}&\text{We cannot simply plug in the value of }x=7\text{, as that would}\\&\text{create a zero value in the denominator, and not tell us much as is.}\\&\text{However, note that the expression}\\&-\frac{(183x + 12x^{2} - 3x^{3} - 840)}{(19x - 133)}\\&\text{can be factored. This will allow us to find the limit}\\&\text{taken from the right:}\\&\frac{(3\cdot (x - 7)\cdot (x - 5)\cdot (x + 8))}{(19\cdot (x - 7))}\\&\frac{(3\cdot (x - 5)\cdot (x + 8))}{19}\\&lim_{x\rightarrow7+}-\frac{(183x + 12x^{2} - 3x^{3} - 840)}{(19x - 133)}=\frac{90}{19}\end{align*}$ $\displaystyle \begin{align*}&\text{We cannot simply plug in the value of }x=7\text{, as that would}\\&\text{create a zero value in the denominator, and not tell us much as is.}\\&\text{However, note that the expression}\\&-\frac{(183x + 12x^{2} - 3x^{3} - 840)}{(19x - 133)}\\&\text{can be factored. This will allow us to find the limit}\\&\text{taken from the right:}\\&\frac{(3\cdot (x - 7)\cdot (x - 5)\cdot (x + 8))}{(19\cdot (x - 7))}\\&\frac{(3\cdot (x - 5)\cdot (x + 8))}{19}\\&lim_{x\rightarrow7+}-\frac{(183x + 12x^{2} - 3x^{3} - 840)}{(19x - 133)}=\frac{90}{19}\end{align*}$

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Example Question #42 : Partial Derivatives

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow7+}\frac{(6x^{2} - 294)}{(13x - 91)}\end{align*}$ $\displaystyle \begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow7+}\frac{(6x^{2} - 294)}{(13x - 91)}\end{align*}$

Possible Answers:

$\infty$ $\displaystyle \infty$

$\frac{84}{13}$ $\displaystyle \frac{84}{13}$

$-\frac{84}{13}$ $\displaystyle -\frac{84}{13}$

$-\infty$ $\displaystyle -\infty$

Correct answer:

$\frac{84}{13}$ $\displaystyle \frac{84}{13}$

Explanation:

$\begin{align*}&\text{We cannot simply plug in the value of }x=7\text{, as that would}\\&\text{create a zero value in the denominator, and not tell us much as is.}\\&\text{However, note that the expression}\\&\frac{(6x^{2} - 294)}{(13x - 91)}\\&\text{can be factored. This will allow us to find the limit}\\&\text{taken from the right:}\\&\frac{(6\cdot (x - 7)\cdot (x + 7))}{(13\cdot (x - 7))}\\&\frac{(6\cdot (x + 7))}{13}\\&lim_{x\rightarrow7+}\frac{(6x^{2} - 294)}{(13x - 91)}=\frac{84}{13}\end{align*}$ $\displaystyle \begin{align*}&\text{We cannot simply plug in the value of }x=7\text{, as that would}\\&\text{create a zero value in the denominator, and not tell us much as is.}\\&\text{However, note that the expression}\\&\frac{(6x^{2} - 294)}{(13x - 91)}\\&\text{can be factored. This will allow us to find the limit}\\&\text{taken from the right:}\\&\frac{(6\cdot (x - 7)\cdot (x + 7))}{(13\cdot (x - 7))}\\&\frac{(6\cdot (x + 7))}{13}\\&lim_{x\rightarrow7+}\frac{(6x^{2} - 294)}{(13x - 91)}=\frac{84}{13}\end{align*}$

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Example Question #42 : Partial Derivatives

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow10-}\frac{(637x + 130x^{2} - 13x^{3} - 6370)}{(6x - 6x^{2} + 540)}\end{align*}$ $\displaystyle \begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow10-}\frac{(637x + 130x^{2} - 13x^{3} - 6370)}{(6x - 6x^{2} + 540)}\end{align*}$

Possible Answers:

$\frac{221}{38}$ $\displaystyle \frac{221}{38}$

$\infty$ $\displaystyle \infty$

$\displaystyle 0$

$-\infty$ $\displaystyle -\infty$

Correct answer:

$\frac{221}{38}$ $\displaystyle \frac{221}{38}$

Explanation:

$\begin{align*}&\text{We cannot simply plug in the value of }x=10\text{, as that would}\\&\text{create a zero value in the denominator, and not tell us much as is.}\\&\text{However, note that the expression}\\&\frac{(637x + 130x^{2} - 13x^{3} - 6370)}{(6x - 6x^{2} + 540)}\\&\text{can be factored. This will allow us to find the limit}\\&\text{taken from the left:}\\&\frac{(13\cdot (x - 10)\cdot (x - 7)\cdot (x + 7))}{(6\cdot (x - 10)\cdot (x + 9))}\\&\frac{(13\cdot (x - 7)\cdot (x + 7))}{(6\cdot (x + 9))}\\&lim_{x\rightarrow10-}\frac{(637x + 130x^{2} - 13x^{3} - 6370)}{(6x - 6x^{2} + 540)}=\frac{221}{38}\end{align*}$ $\displaystyle \begin{align*}&\text{We cannot simply plug in the value of }x=10\text{, as that would}\\&\text{create a zero value in the denominator, and not tell us much as is.}\\&\text{However, note that the expression}\\&\frac{(637x + 130x^{2} - 13x^{3} - 6370)}{(6x - 6x^{2} + 540)}\\&\text{can be factored. This will allow us to find the limit}\\&\text{taken from the left:}\\&\frac{(13\cdot (x - 10)\cdot (x - 7)\cdot (x + 7))}{(6\cdot (x - 10)\cdot (x + 9))}\\&\frac{(13\cdot (x - 7)\cdot (x + 7))}{(6\cdot (x + 9))}\\&lim_{x\rightarrow10-}\frac{(637x + 130x^{2} - 13x^{3} - 6370)}{(6x - 6x^{2} + 540)}=\frac{221}{38}\end{align*}$

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Example Question #49 : Partial Derivatives

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow3+}\frac{(5x^{2} - 50x + 105)}{(3x^{2} - 18x + 27)}\end{align*}$ $\displaystyle \begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow3+}\frac{(5x^{2} - 50x + 105)}{(3x^{2} - 18x + 27)}\end{align*}$

Possible Answers:

$\infty$ $\displaystyle \infty$

$-\infty$ $\displaystyle -\infty$

$\displaystyle 0$

$\frac{ 1}{5}$ $\displaystyle \frac{ 1}{5}$

Correct answer:

$-\infty$ $\displaystyle -\infty$

Explanation:

$\begin{align*}&\text{We cannot simply plug in the value of }x=3\text{, as that would}\\&\text{create a zero value in the denominator, and not tell us much as is.}\\&\text{However, note that the expression}\\&\frac{(5x^{2} - 50x + 105)}{(3x^{2} - 18x + 27)}\\&\text{can be factored. This will allow us to find the limit}\\&\text{taken from the right:}\\&\frac{(5\cdot (x - 3)\cdot (x - 7))}{(3\cdot (x - 3)^{2})}\\&\frac{(5\cdot (x - 7))}{(3\cdot (x - 3))}\\&lim_{x\rightarrow3+}\frac{(5x^{2} - 50x + 105)}{(3x^{2} - 18x + 27)}=-\infty\end{align*}$ $\displaystyle \begin{align*}&\text{We cannot simply plug in the value of }x=3\text{, as that would}\\&\text{create a zero value in the denominator, and not tell us much as is.}\\&\text{However, note that the expression}\\&\frac{(5x^{2} - 50x + 105)}{(3x^{2} - 18x + 27)}\\&\text{can be factored. This will allow us to find the limit}\\&\text{taken from the right:}\\&\frac{(5\cdot (x - 3)\cdot (x - 7))}{(3\cdot (x - 3)^{2})}\\&\frac{(5\cdot (x - 7))}{(3\cdot (x - 3))}\\&lim_{x\rightarrow3+}\frac{(5x^{2} - 50x + 105)}{(3x^{2} - 18x + 27)}=-\infty\end{align*}$

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Example Question #44 : Limits

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow-2+}-\frac{(792x + 180x^{2} + 12x^{3} + 960)}{(128x + 8x^{2} - 8x^{3} + 160)}\end{align*}$ $\displaystyle \begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow-2+}-\frac{(792x + 180x^{2} + 12x^{3} + 960)}{(128x + 8x^{2} - 8x^{3} + 160)}\end{align*}$

Possible Answers:

$\displaystyle 0$

$-\infty$ $\displaystyle -\infty$

$\frac{ 8}{5}$ $\displaystyle \frac{ 8}{5}$

$\infty$ $\displaystyle \infty$

Correct answer:

$-\infty$ $\displaystyle -\infty$

Explanation:

$\begin{align*}&\text{We cannot simply plug in the value of }x=-2\text{, as that would}\\&\text{create a zero value in the denominator, and not tell us much as is.}\\&\text{However, note that the expression}\\&-\frac{(792x + 180x^{2} + 12x^{3} + 960)}{(128x + 8x^{2} - 8x^{3} + 160)}\\&\text{can be factored. This will allow us to find the limit}\\&\text{taken from the right:}\\&\frac{(12\cdot (x + 2)\cdot (x + 5)\cdot (x + 8))}{(8\cdot (x + 2)^{2}\cdot (x - 5))}\\&\frac{(3\cdot (x + 5)\cdot (x + 8))}{(2\cdot (x + 2)\cdot (x - 5))}\\&lim_{x\rightarrow-2+}-\frac{(792x + 180x^{2} + 12x^{3} + 960)}{(128x + 8x^{2} - 8x^{3} + 160)}=-\infty\end{align*}$ $\displaystyle \begin{align*}&\text{We cannot simply plug in the value of }x=-2\text{, as that would}\\&\text{create a zero value in the denominator, and not tell us much as is.}\\&\text{However, note that the expression}\\&-\frac{(792x + 180x^{2} + 12x^{3} + 960)}{(128x + 8x^{2} - 8x^{3} + 160)}\\&\text{can be factored. This will allow us to find the limit}\\&\text{taken from the right:}\\&\frac{(12\cdot (x + 2)\cdot (x + 5)\cdot (x + 8))}{(8\cdot (x + 2)^{2}\cdot (x - 5))}\\&\frac{(3\cdot (x + 5)\cdot (x + 8))}{(2\cdot (x + 2)\cdot (x - 5))}\\&lim_{x\rightarrow-2+}-\frac{(792x + 180x^{2} + 12x^{3} + 960)}{(128x + 8x^{2} - 8x^{3} + 160)}=-\infty\end{align*}$

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Calculus 3 : Partial Derivatives

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #41 : Partial Derivatives

Example Question #41 : Limits

Example Question #41 : Limits

Example Question #1002 : Calculus 3

Example Question #1001 : Calculus 3

Example Question #43 : Partial Derivatives

Example Question #42 : Partial Derivatives

Example Question #42 : Partial Derivatives

Example Question #49 : Partial Derivatives

Example Question #44 : Limits

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